-0.000 000 000 742 147 676 602 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 602(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 602(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 602| = 0.000 000 000 742 147 676 602


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 602.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 602 × 2 = 0 + 0.000 000 001 484 295 353 204;
  • 2) 0.000 000 001 484 295 353 204 × 2 = 0 + 0.000 000 002 968 590 706 408;
  • 3) 0.000 000 002 968 590 706 408 × 2 = 0 + 0.000 000 005 937 181 412 816;
  • 4) 0.000 000 005 937 181 412 816 × 2 = 0 + 0.000 000 011 874 362 825 632;
  • 5) 0.000 000 011 874 362 825 632 × 2 = 0 + 0.000 000 023 748 725 651 264;
  • 6) 0.000 000 023 748 725 651 264 × 2 = 0 + 0.000 000 047 497 451 302 528;
  • 7) 0.000 000 047 497 451 302 528 × 2 = 0 + 0.000 000 094 994 902 605 056;
  • 8) 0.000 000 094 994 902 605 056 × 2 = 0 + 0.000 000 189 989 805 210 112;
  • 9) 0.000 000 189 989 805 210 112 × 2 = 0 + 0.000 000 379 979 610 420 224;
  • 10) 0.000 000 379 979 610 420 224 × 2 = 0 + 0.000 000 759 959 220 840 448;
  • 11) 0.000 000 759 959 220 840 448 × 2 = 0 + 0.000 001 519 918 441 680 896;
  • 12) 0.000 001 519 918 441 680 896 × 2 = 0 + 0.000 003 039 836 883 361 792;
  • 13) 0.000 003 039 836 883 361 792 × 2 = 0 + 0.000 006 079 673 766 723 584;
  • 14) 0.000 006 079 673 766 723 584 × 2 = 0 + 0.000 012 159 347 533 447 168;
  • 15) 0.000 012 159 347 533 447 168 × 2 = 0 + 0.000 024 318 695 066 894 336;
  • 16) 0.000 024 318 695 066 894 336 × 2 = 0 + 0.000 048 637 390 133 788 672;
  • 17) 0.000 048 637 390 133 788 672 × 2 = 0 + 0.000 097 274 780 267 577 344;
  • 18) 0.000 097 274 780 267 577 344 × 2 = 0 + 0.000 194 549 560 535 154 688;
  • 19) 0.000 194 549 560 535 154 688 × 2 = 0 + 0.000 389 099 121 070 309 376;
  • 20) 0.000 389 099 121 070 309 376 × 2 = 0 + 0.000 778 198 242 140 618 752;
  • 21) 0.000 778 198 242 140 618 752 × 2 = 0 + 0.001 556 396 484 281 237 504;
  • 22) 0.001 556 396 484 281 237 504 × 2 = 0 + 0.003 112 792 968 562 475 008;
  • 23) 0.003 112 792 968 562 475 008 × 2 = 0 + 0.006 225 585 937 124 950 016;
  • 24) 0.006 225 585 937 124 950 016 × 2 = 0 + 0.012 451 171 874 249 900 032;
  • 25) 0.012 451 171 874 249 900 032 × 2 = 0 + 0.024 902 343 748 499 800 064;
  • 26) 0.024 902 343 748 499 800 064 × 2 = 0 + 0.049 804 687 496 999 600 128;
  • 27) 0.049 804 687 496 999 600 128 × 2 = 0 + 0.099 609 374 993 999 200 256;
  • 28) 0.099 609 374 993 999 200 256 × 2 = 0 + 0.199 218 749 987 998 400 512;
  • 29) 0.199 218 749 987 998 400 512 × 2 = 0 + 0.398 437 499 975 996 801 024;
  • 30) 0.398 437 499 975 996 801 024 × 2 = 0 + 0.796 874 999 951 993 602 048;
  • 31) 0.796 874 999 951 993 602 048 × 2 = 1 + 0.593 749 999 903 987 204 096;
  • 32) 0.593 749 999 903 987 204 096 × 2 = 1 + 0.187 499 999 807 974 408 192;
  • 33) 0.187 499 999 807 974 408 192 × 2 = 0 + 0.374 999 999 615 948 816 384;
  • 34) 0.374 999 999 615 948 816 384 × 2 = 0 + 0.749 999 999 231 897 632 768;
  • 35) 0.749 999 999 231 897 632 768 × 2 = 1 + 0.499 999 998 463 795 265 536;
  • 36) 0.499 999 998 463 795 265 536 × 2 = 0 + 0.999 999 996 927 590 531 072;
  • 37) 0.999 999 996 927 590 531 072 × 2 = 1 + 0.999 999 993 855 181 062 144;
  • 38) 0.999 999 993 855 181 062 144 × 2 = 1 + 0.999 999 987 710 362 124 288;
  • 39) 0.999 999 987 710 362 124 288 × 2 = 1 + 0.999 999 975 420 724 248 576;
  • 40) 0.999 999 975 420 724 248 576 × 2 = 1 + 0.999 999 950 841 448 497 152;
  • 41) 0.999 999 950 841 448 497 152 × 2 = 1 + 0.999 999 901 682 896 994 304;
  • 42) 0.999 999 901 682 896 994 304 × 2 = 1 + 0.999 999 803 365 793 988 608;
  • 43) 0.999 999 803 365 793 988 608 × 2 = 1 + 0.999 999 606 731 587 977 216;
  • 44) 0.999 999 606 731 587 977 216 × 2 = 1 + 0.999 999 213 463 175 954 432;
  • 45) 0.999 999 213 463 175 954 432 × 2 = 1 + 0.999 998 426 926 351 908 864;
  • 46) 0.999 998 426 926 351 908 864 × 2 = 1 + 0.999 996 853 852 703 817 728;
  • 47) 0.999 996 853 852 703 817 728 × 2 = 1 + 0.999 993 707 705 407 635 456;
  • 48) 0.999 993 707 705 407 635 456 × 2 = 1 + 0.999 987 415 410 815 270 912;
  • 49) 0.999 987 415 410 815 270 912 × 2 = 1 + 0.999 974 830 821 630 541 824;
  • 50) 0.999 974 830 821 630 541 824 × 2 = 1 + 0.999 949 661 643 261 083 648;
  • 51) 0.999 949 661 643 261 083 648 × 2 = 1 + 0.999 899 323 286 522 167 296;
  • 52) 0.999 899 323 286 522 167 296 × 2 = 1 + 0.999 798 646 573 044 334 592;
  • 53) 0.999 798 646 573 044 334 592 × 2 = 1 + 0.999 597 293 146 088 669 184;
  • 54) 0.999 597 293 146 088 669 184 × 2 = 1 + 0.999 194 586 292 177 338 368;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 602(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 602(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 602(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 602 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111